0-1 backpack

1. poj 3211 inserting clothes

Analysis: the order of clothes in a certain color is arbitrary, and the problem can be converted to: When the clothes in a certain color are washed at least, two people can wash them at the same time, to minimize the time difference between two people, that isTake 1/2 of the total time as the backpack capacity, and find the value closest to 1/2.

0-1 backpack with only weight and no weight:

Memset (F,-1, sizeof (f); // Initialization
F [0] = 0;
For (j = 0; j <CT [I]; j ++ ){
For (int K = vol-time [I] [J]; k> = 0; k --){
If (F [k]> = 0 ){
F [K + time [I] [J] = 1;
}
}
}

View code

 1 # Include <stdio. h> 2 # Include <String . H> 3   4   Int  Main ()  5   {  6       Int  N, m;  7       Char Clor [ 15 ] [ 15  ];  8       Int Time [15 ] [ 100 ], CT [ 15 ], Sum [ 15  ];  9       Int F [ 100010  ];  10       While (Scanf ( "  % D  " , & N, & M) & (N | M )){ 11           Int  I, J;  12   13           For (I = 0 ; I <n; I ++ )  14 Scanf ( "  % S  "  , Clor [I]);  15 Memset (CT, 0 , Sizeof (CT ));  16 Memset (sum, 0 , Sizeof  (SUM ));  17           For (I = 0 ; I <m; I ++ ){  18               Int  T;  19               Char Cl [ 15  ]; 20   21 Scanf ( "  % D % s  " ,& T, Cl );  22               For (J = 0 ; J <n; j ++ ){  23                   If (Strcmp (CL, clor [J]) = 0  ){  24 Time [J] [cT [J] ++] =T;  25 Sum [J] + = T;  26                       Break  ;  27   }  28   }  29   }  30           31           Int St = 0 ;  32           For (I = 0 ; I <n; I ++ ){  33               Int Vol = sum [I]/ 2  ;  34 Memset (F ,- 1 , Sizeof  (F ));  35 F [ 0 ] =0  ;  36               For (J = 0 ; J <CT [I]; j ++ ){  37                   For ( Int K = vol-time [I] [J]; k> = 0 ; K -- ){  38                       If (F [k]> = 0  ){ 39 F [K + time [I] [J] = 1  ;  40   }  41   }  42   }  43               Int V1 = Vol;  44               While (F [V1] < 0 ) V1 --;  45 St + = sum [I]- V1;  46               47   }  48 Printf ( "  % D \ n  "  , St );  49   }  50       Return   0  ; 51 }